Given a set of vectors
S = $\left\{
\begin{bmatrix} 2 \\ 1 \end{bmatrix},
\begin{bmatrix} 3 \\ 2 \end{bmatrix},
\begin{bmatrix} 1 \\ 2 \end{bmatrix}
\right\}
$
Find out if the vectors are linearly dependent or independent
I know that for a set of vectors to be linearly dependent, they must satisfy the below equation:
$$c_1v_1 + c_2v_2 … c_nv_n = \mathbf 0 $$
such that not all $c_i$ are zero.
So, I decided to apply Gauss Elimination and I got the following equation:
$
c_1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 3 \\ 2 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}
$
And, needless to say, I get an under-determined system of equations below:
$$2c_1 + 3c_2 + c3 = 0$$
$$c_1 + 2c_2 + 2c_3 = 0$$
And after solving, I get this:
$$c_1 = 4c_3$$
$$c_2 = -3c_3$$
So, $c_3$ is the free variable.
Assuming $c_3$ is non zero, the vectors are linearly dependent. If it is zero, vectors are linearly dependent.
How can it be that a free variable decides whether vectors are linearly dependent or not ? Shouldn't it a 100% yes or no answer that does not fluctuate depending on values of constants?
Best Answer
The answer does not fluctuate; you’ve misunderstood the definition of linear independence. A set $\{v_1,\dots,v_n\}$ of vectors is linearly dependent if and only if there is at least one set of coefficients $\{c_1,\dots,c_n\}$ such that
$$c_1v_1+\ldots c_nv_n=0$$
and at least one of the coefficients is non-zero. It is linearly independent if and only if it is not linearly dependent. Your calculations show that $4v_1-3v_2+v_3=0$, with coefficients $4,-3$, and $1$; it’s certainly true that at least one of these is non-zero(!), so your set of vectors is linearly dependent.
Note that it’s always true that $0v_1+\ldots 0v_n=0$; that’s never at issue and does not make the set of vectors linearly dependent. The question is whether there is any other set of coefficients that makes the sum $0$. If there is, the set of vectors is linearly dependent; if not, it’s linearly independent.